# Rank and nullity of linear maps

1. May 9, 2009

### Jennifer1990

1. The problem statement, all variables and given/known data
By considering the dimensions of the range or null space, determine the rank and the nullity of the following linear maps:
a) D:Pn --> Pn-1, where D(x^k)=Kx^k-1
b) L:M(2,3) --> M(2,3) where L([a b c; d e f])=[d e f; 0 0 0]
c) Tr:M(3,3) --> R, where Tr(A)=a11+a22+a33 (the trace of A)

2. Relevant equations
None

3. The attempt at a solution
a)If we rewrite Pn-1 as a column vector and row-reduce, we are left with just 1 in the first row and all the other entries are zeroes. Therefore the nullity is 1 since we have one parameter in the solution space. The rank is 1 - nullity = 0

b)Solving for the nullspace we see that there are 2 parameters in the solution space therefore the nullity is 2 and the rank is 1

c)We have one equation in 3 unknown numbers .The kernel consists of all a11, a22, a33 such that a11+a22+a33 = 0. We can solve that for one of the variables in terms of the other 2. Therefore the nullity is 2 and the rank is 1.
Can you give me a matrix which represents the linear mapping Tr? :S

I am not sure if I did these correctly. I would appreciate any feedback possible. thank you!

2. May 9, 2009

### jbunniii

With regard to (a), I assume Pn denotes the space of polynomials of degree $$\leq n$$ and whose coefficients are elements of some ring or field. Therefore D is the derivative operator. Which subspace of polynomials have derivative equal to zero, and what is the dimension of this subspace? Then, in general, what can you say about

dim(range(A)) + dim(null(A))

where A is any linear map between finite-dimensional vector spaces?

With regard to (b), tell us what you have determined the null space of L to be. That is, if you apply L to a general matrix M = [a b c; d e f], what are the necessary and sufficient conditions on a, b, c, d, e, and f for M to be in the null space of L?

With regard to (c), since T maps a 3x3 matrix to a real number, you should be writing "the null space of T consists of those 3x3 matrices for which..." It does NOT consist of specific "a11, a22, a33" which are real numbers, not matrices.

3. May 9, 2009

### Jennifer1990

a) The subspace of polynomials that have derivative equal to zero are those with x^0. So then this means that the rank is 1 and the nullity is 0 by the rank-nullity theorem?

b) Applying L to a general matrix M = [a b c; d e f] gives [d e f; 0 0 0]. For M to be the nullspace of L, then d e and f must also equal 0. Solving for the nullspace, we get x3 = 0, x2 = 0. These are free variables which are always equal to 0 therefore for kd + e + f = 0, k is a real number, d must equal 0. Therefore, the nullity is 2 and the rank is 1

c)"the null space of T consists of those 3x3 matrices for which a11+a22+a33 = 0"
I think this is what you are asking for. And I still think the answer is nullity = 2 and rank =1...

Could you let me know if the answers I gave are right? thank you~

4. May 9, 2009

### jbunniii

I guess you mean to say, the subspace of polynomials that have derivative equal to zero are those with degree 0, i.e., the ones that don't have any powers of x higher than x^0. OK, that's correct.

Now, what does "nullity" mean? It means the dimension of the null space. You just described what the null space is: it is the space of polynomials of degree 0. What is the dimension of this space? Hint: it's not zero. It's the answer to the following question. Given an arbitrary element of Pn, which looks like,

$$a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x^1 + a_0 x^0[\tex], how many of the coefficients are constrained by the definition of the null space of D? Your answer for the rank is also incorrect, but we'll get to that after you work out the nullity. OK, good. No, this isn't right. I don't think you're understanding exactly what "nullspace" means. The null space of L is precisely the set of 2x3 matrices [a b c; d e f] for which d = e = f = 0. The dimension of this null space is once again the answer to this question: How many of the parameters a, b, c, d, e, f are constrained by the definition of the null space of L? The rank is also incorrect, but let's get to that after you work out the dimension of the null space. OK, that's correct. Your answer for the nullity is correct, but I'm not 100% sure that your reasoning is correct. An arbitrary 3x3 matrix has NINE coefficients. How many of these are constrained by the condition a11+a22+a33 = 0? Also, rank = 1 is incorrect. Can you tell me the definition of the rank of a linear map, say from a vector space V to another vector space W? (Hint: it's the dimension of a certain subspace associated with the map. Which subspace, and what space is it a subspace of: V or W?) Can you tell me what this subspace is in each of the three examples above? Also, can you tell me what you know about dim(null(A)) + rank(A) for a given linear map between two finite-dimensional vector spaces? Last edited: May 9, 2009 5. May 10, 2009 ### Jennifer1990 a)Given an arbitrary element of Pn, which looks like, [tex]a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x^1 + a_0 x^0[\tex], how many of the coefficients are constrained by the definition of the null space of D? One of the coefficients is constrained by the definition of the nullspace of D. It is the coefficient of x^0. Therefore, the nullity is 1 In Pn-1, there are n different polynomials, polynomials of degree 0, 1, 2,...n-1 Therefore, the rank = n-1......i think... b)How many of the parameters a, b, c, d, e, f are constrained by the definition of the null space of L? 3 of the parameters are constrained by the definition of the null space Therefore, the nullity is 3 Can I see how you would solve for the nullspace? By rank-nullity theorem, rank = 0 c)Can you tell me the definition of the rank of a linear map, say from a vector space V to another vector space W? The rank of a linear map, from vector space V to vector space W is the dimension of the range space of V, which is a subspace of W dim(null(A)) + rank(A) for a given linear map between two finite-dimensional vector spaces dim(null(A)) + rank(A) = number of columns in matrix A (rank-nullity theorem) 6. May 10, 2009 ### jbunniii No, if that were true then I would be free to pick the coefficients of x^1 through x^n anyway I like, yet the polynomial would be in the null space of D? i.e., its derivative would equal zero? Let me rephrase the question: if the polynomial [tex]a_n x^n + a_{n-1}x^{n-1} + \ldots + a_1 x^1 + a_0$$

has derivative equal to zero, then what does that say about the coefficients? Which coefficients are forced to be zero, and which can you choose any way you like?

L maps [a b c; d e f] to [d e f; 0 0 0]. The definition of the null space is the set of those matrices [a b c; d e f] that L maps to zero. That is, it is the set of all matrices [a b c; d e f] with d = e = f = 0. Therefore if asked "what is the null space?" you could say: "it is the set of all matrices of the form [a b c; 0 0 0]" where a, b, c are chosen freely from whichever field the coefficients come from (real or complex numbers, typically).

L maps from V to V, where V is the space of 2x3 matrices. What is the dimension of V? Hint: it is not 2, and it is not 3. How many degrees of freedom do you have when you choose a 2x3 matrix? That is the dimension of V.

Then, the rank-nullity theorem tells us that

dim(null L) + rank(L) = dim(V)

If you know dim(null L) is 3 and you know dim(V) by answering the question above, then rank(L) is dim(V) - dim(null L).

7. May 11, 2009

### Jennifer1990

a)All coefficients other than the coefficient of x^0 must be equal to 0 in order for the polynomial to have a derivative of 0. Therefore, n coefficients are constrained by the definition of the null space and we can choose one coefficient any way we like (i.e. the coefficient of x^0)
Therefore, the nullspace is n
Can you give me a hint on how to find the rank?

b)You said that the dimension of V is not 3. How can this be?
Doesn't the range space of L have matrices of 2 rows? So, the nullity of L which is 3 plus the dimension of V must equal 2?

8. May 11, 2009

### jbunniii

Everything you said is correct except for your conclusion. We have one degree of freedom in choosing coefficients to make the derivative zero. Therefore the dimension of the null space is 1.

$$dim(null(D)) + rank(D) = dim(P^{n-1})$$

First, understand why this is true. In particular, why is the right-hand side $$dim(P^{n-1})$$ and not $$dim(P^n)$$? Then, what is $$dim(P^{n-1})$$? Then, just solve for rank(D).

How many degrees of freedom do I have to choose a 2x3 matrix? There are six elements in such a matrix, and I can choose them any way I like, right? Therefore the dimension of V is six. Does this make sense?

9. May 11, 2009

### Jennifer1990

a)Don't you mean to say $$dim(null(D)) + rank(D) = dim(P^{n})$$
Because $$P^{n-1}$$ refers to the range space of D....
The rank nullity-theorem states $$dim(null(D)) + rank(D) = dim(D)$$ and in this case the dimension of D is Pn

b) I think I see what you are tying to say. For the matrix A=[d e f; 0 0 0] to be in the nullspace, d=e=f=0. You have 6 elements and can choose in any way you like. Therefore, the nullity is 6-3(i.e. the number of elements which are not constrained by the nullspace). The rank is also 3

10. May 11, 2009

### jbunniii

Yes, I meant Pn. But it's not correct to say that the dimension of D is Pn. The dimension of D is a number. Pn is a space. You meant to say that the dimension of D is the dimension of Pn. So what is the dimension of Pn? (This means: how many degrees of freedom do you have when choosing an element of Pn?)

Correct.

11. May 11, 2009

### Jennifer1990

a) we have n elements which we can choose in any way we like

b)I am wondering how the rank of the matrix can possibly be 3 because in a 2x3 matrix, the highest rank it can have is 2...right?

12. May 11, 2009

### jbunniii

An element of $$P^n$$ looks like

$$a_n x^n + a_{n-1}x^{n-1} + \ldots + a_1 x + a_0$$

How many a's are there? (You're off by one!)

You're confusing the matrices mapped by L with the matrix of the linear map L.

L maps one 2x3 matrix into another 2x3 matrix. We haven't actually expressed L itself as a matrix, but if we did, it would have to be 6x6! (And you would have to take the elements of the 2x3 input and output matrices and stretch them out into 6x1 vectors.)

13. May 11, 2009

### Jennifer1990

a)There are n+1 a's?

14. May 11, 2009

### jbunniii

Yes, so that's the dimension of Pn.